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  • Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise multiple choice question 7

Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise multiple choice question 7

Answers (1)

Answer:

 Option (a)

Hint:

 Use cross product of vector.

Given:

 \vec{r}=\hat{i}-\hat{j}+\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (\hat{i}-2\hat{j}+3\hat{k})

Solution:

The given plane is

\vec{r}=\hat{i}-\hat{j}+\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (\hat{i}-2\hat{j}+3\hat{k})

So, it is clear from the given equation of plane, that the plane passing through a point (\hat{i}-\hat{j}) and parallel to the two vectors (\hat{i}+\hat{j}+\hat{k}) \text { and } (\hat{i}-2\hat{j}+3\hat{k}).

Therefore, the equation of the vector normal to the plane is given as

\begin{aligned} &\widehat{n}=[(\hat{i}+\hat{j}+\hat{k}) \times (\hat{i}-2\hat{j}+3\hat{k})]\\ &=\begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1 &1 &1 \\ 1 &-2 &3 \end{vmatrix}\\ &=\left [ \left \{ (1\times 3)-(-2\times 1) \right \}\widehat{i}-\left \{ (1\times 3)-(1\times 1) \right \}\widehat{j}+\left \{ (1\times -2)-(1\times 1)\widehat{k} \right \} \right ]\\ &=5\widehat{i}-2\widehat{j}-3\widehat{k} \end{aligned}

So, in scalar product form the vector equation of the plane is given as,

\begin{aligned} &(\overrightarrow{r}-( \hat{i}-\hat{j})).\widehat{n}=0\\ &(\overrightarrow{r}-( \hat{i}-\hat{j})).(5\widehat{i}-2\widehat{j}-3\widehat{k})=0\\ &\overrightarrow{r}.(5\widehat{i}-2\widehat{j}-3\widehat{k})=5+2\\ &\overrightarrow{r}.(5\widehat{i}-2\widehat{j}-3\widehat{k})=7 \end{aligned}

Posted by

Gurleen Kaur

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